Question

If \[ \int \frac{\tan x}{1+\tan x+\tan^2x}\,dx = x-\frac{K}{\sqrt A} \tan^{-1} \left( \frac{K\tan x+1}{\sqrt A} \right) +C, \] then the ordered pair \((K,A)\) is

• A. \((2,3)\)
• B. \((2,1)\)
• C. \((-2,1)\)
• D. \((-2,3)\)

Hint:

For integrals involving \(\tan x\), use \(t=\tan x\). After converting to a rational function, complete the square in the denominator to obtain inverse tangent terms.

Answer 1

The Correct Option is A

Concept: Convert the integral into a rational function by putting \[ t=\tan x. \] Then use partial fractions and compare with the given result.

Step 1: Substitute \(t=\tan x\). Then \[ dt=\sec^2x\,dx=(1+t^2)\,dx, \] so that \[ dx=\frac{dt}{1+t^2}. \] Hence, \[ I = \int \frac{t}{(1+t+t^2)(1+t^2)} \,dt. \]

Step 2: Resolve into partial fractions. Assume \[ \frac{t}{(1+t+t^2)(1+t^2)} = \frac{At+B}{1+t+t^2} + \frac{Ct+D}{1+t^2}. \] Multiplying throughout by \[ (1+t+t^2)(1+t^2), \] \[ t = (At+B)(1+t^2) + (Ct+D)(1+t+t^2). \] Comparing coefficients gives \[ A=-1,\qquad B=0, \qquad C=1,\qquad D=1. \] Thus, \[ \frac{t}{(1+t+t^2)(1+t^2)} = -\frac{t}{1+t+t^2} + \frac{t+1}{1+t+t^2}. \] Therefore, \[ I = \int\frac{t+1}{1+t+t^2}\,dt -\int\frac{t}{1+t^2}\,dt. \]

Step 3: Integrate. Write \[ t+1 = \frac12(2t+1)+\frac12. \] Hence, \[ \int\frac{t+1}{t^2+t+1}\,dt = \frac12\ln(t^2+t+1) +\frac12\int\frac{dt}{t^2+t+1}. \] Now, \[ t^2+t+1 = \left(t+\frac12\right)^2+\frac34. \] Therefore, \[ \int\frac{dt}{t^2+t+1} = \frac{2}{\sqrt3} \tan^{-1} \left( \frac{2t+1}{\sqrt3} \right). \] Also, \[ \int\frac{t}{1+t^2}\,dt = \frac12\ln(1+t^2). \] Combining and simplifying, \[ I = x - \frac{2}{\sqrt3} \tan^{-1} \left( \frac{2\tan x+1}{\sqrt3} \right) +C. \]

Step 4: Compare with the given expression. Given, \[ I = x - \frac{K}{\sqrt A} \tan^{-1} \left( \frac{K\tan x+1}{\sqrt A} \right) +C. \] Comparing, \[ K=2, \qquad A=3. \] Therefore, \[ \boxed{(K,A)=(2,3)} \] \[ \boxed{\text{Answer = (A)}} \]